right=0,1# left=buy, right=sellmaxP=0whileright<len(prices):## 遍历整个 listifprices[right]>prices[left]:## 在存在赚钱机会的条件下profit=prices[right]-prices[left]maxP=max(maxP,profit)else:## 对于任意一个右指针日期,我们都要回顾它后面的所有日期,
【leetcode】43-best-time-to-buy-and-sell-stock-iv 力扣 188. 买卖股票的最佳时机 IV 【leetcode】44-best-time-to-buy-and-sell-stock-with-cooldown 力扣 309. 买卖股票的最佳时机包含冷冻期 【leetcode】45-best-time-to-buy-and-sell-stock-with-cooldown 力扣 714. 买卖股票的最佳时机包含手续费 ...
【leetcode】43-best-time-to-buy-and-sell-stock-iv 力扣 188. 买卖股票的最佳时机 IV 【leetcode】44-best-time-to-buy-and-sell-stock-with-cooldown 力扣 309. 买卖股票的最佳时机包含冷冻期 【leetcode】45-best-time-to-buy-and-sell-stock-with-cooldown 力扣 714. 买卖股票的最佳时机包含手续费 ...
解法: classSolution{public:intmaxProfit(vector<int>& prices){intsell =0;intbuy =0;if(prices.empty()){return0; }else{ buy = -prices[0]; }for(intprice : prices){ sell =max(sell, buy + price); buy =max(buy, -price); }returnsell; } };...
第一个:https://leetcode.com/problems/best-time-to-buy-and-sell-stock/?tab=Description 至多交易一次,返回最大利润。实质上是让求一个连续的子串,其最后一个元素和第一个元素只差最大。因为题目要求至多一次交易而不是必须一次,因此如果数组降序,可以选择不交易,所以理论上最小值是0,对应不...
Can you solve this real interview question? Best Time to Buy and Sell Stock - You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choo
问题https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/ 练习使用JavaScript解答 /** * @param {number[]} prices * @return {number} */ var maxProfit = function(prices) { if(prices.length < 2) return 0; ...
Can you solve this real interview question? Best Time to Buy and Sell Stock III - You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transacti
LeetCode 0121. Best Time to Buy and Sell Stock买卖股票的最佳时机【Easy】【Python】【贪心】 Problem LeetCode Say you have an array for which theith element is the price of a given stock on dayi. If you were only permitted to complete at most one transaction (i.e., buy one and sell...
最终的结果为sell2, 因为sell2包含了一次交易和二次交易的最大收益class Solution { public: int maxProfit(vector<int>& prices) { int n = prices.size(); int dp[n][2]; //vector 更耗时 memset(dp, 0, sizeof(dp)); int minv = prices[0], maxv = prices[n - 1], ans = 0...